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Subnetting question that has me stuck

Discussion in 'Routing & Switching' started by BraderzTheDog, Apr 20, 2012.

  1. BraderzTheDog

    BraderzTheDog Kilobyte Poster

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    Hola,

    I've been studying frantically lately to sit my CCENT exam this Tuesday, however I'm struggling with 1 type of subnetting question :eek:

    So far I can reverse engineer an IP address find its subnet mask and increment and in under a minute i've found its network range and know the network / broadcast and usable host addresses. (Not any concern there)

    However its one particular style of question im having difficulty with. Ill paste an example below and tell you my though process and then where i hit the wall. Im sure you can help :)

    How many subnets and hosts per subnet can you get from the network 172.29.0.0/23?

    Well I need to know the maximum for each the networks and the hosts possible, so I would go about completing the SNM 255.255.254.0 with an increment of 2. I can get the ranges however there are gonna be so many ranges with such a small increment how am I meant to figure out the number of hosts / networks available without writing them all down etc...

    172.29.0.0 - 172.29.1.255
    172.29.2.0 - 172.29.3.255
    172.29.4.0 - 172.29.5.255

    etc...

    Help! Surely there is a quicker method, or am i just doing it wrong as it were :rolleyes:

    All help much appreciated!

    Kind Regards,
    Brad.
     
    Certifications: CCNA R&S, CCNA-SEC, CCSA, JNCIA FWV, MCITP, MCTS, MTA, A+
  2. danielno8

    danielno8 Gigabyte Poster

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    128 subnets, 510 hosts.

    I'll not be able to explain it well, but basically 7 bits are used for the subnet portion (23-16) and 9 bits (32-23) used for the host portion. Remeber to minus two from the numebr you calculate for the host for the subnet/broadcast address.

    Have a read through subnetting again and you will see this explained properly!
     
    Last edited: Apr 20, 2012
    Certifications: CCENT, CCNA
    WIP: CCNP
  3. Nyx

    Nyx Byte Poster

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    Remember, subnet mask tells you how many bits in the address is reserved for network address, rest of the bits is for hosts. ie:
    172.29.0.0/23 leaves you with 9 bits for hosts, this gives you 512 addresses. two addresses reserved for network and broadcast address - so 510 hosts max (if you don't subnet!) 2 to the power of bits left - 2 (for the network and broadcast)
    increasing the network mask by one /24 gives you one bit for subnets, by two /25 gives you 4 (two bits) ..... by eight /31 gives you 8 bits - 128 addresses for subnets.

    ranges you wrote down are wrong as you have only one bit to change in the third octect! so first three octets will be 172.29.0 or 172.29.1, you would need to supernet to have more.
     
  4. danielno8

    danielno8 Gigabyte Poster

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    Those are correct subnet ranges....

    172.29.0.0 --> 172.29.1.255, with 172.29.0.0 being reserved for the network, and 172.29.1.255 being reserved for broadcast. All addresses in between are available for hosts.

    172.29.2.0 --> 172.29.3.255 with 172.29.2.0 being reserved for the network, and 172.29.3.255 being reserved for broadcast. All addresses in between are available for hosts.

    172.29.4.0 --> 172.29.5.255 with 172.29.4.0 being reserved for the network, and 172.29.5.255 being reserved for broadcast. All addresses in between are available for hosts.

    etc etc

    EDIT: never mind, i missed that the initial address given was 172.29.0.0/23 rather than 172.29.0.0/16
     
    Last edited: Apr 20, 2012
    Certifications: CCENT, CCNA
    WIP: CCNP
  5. BraderzTheDog

    BraderzTheDog Kilobyte Poster

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    Thanks,

    So you get the 9 bits by filling in the rest of the subnet mask?

    For example that IP address had a /23 leaving 9 bits to make the full address 255.255.255.255. I think that makes sense now, that was the missing part of the puzzle.

    Regards,
    Brad.
     
    Certifications: CCNA R&S, CCNA-SEC, CCSA, JNCIA FWV, MCITP, MCTS, MTA, A+

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